Question: Simplify and expand the following expression: $ \dfrac{2}{n - 8}- \dfrac{3}{3n - 12}- \dfrac{2}{n^2 - 12n + 32} $
Explanation: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $3$ out of denominator in the second term: $ \dfrac{3}{3n - 12} = \dfrac{3}{3(n - 4)}$ We can factor the quadratic in the third term: $ \dfrac{2}{n^2 - 12n + 32} = \dfrac{2}{(n - 8)(n - 4)}$ Now we have: $ \dfrac{2}{n - 8}- \dfrac{3}{3(n - 4)}- \dfrac{2}{(n - 8)(n - 4)} $ The least common multiple of the denominators is: $ (n - 8)(n - 4)$ In order to get the first term over $(n - 8)(n - 4)$ , multiply by $\dfrac{3(n - 4)}{3(n - 4)}$ $ \dfrac{2}{n - 8} \times \dfrac{3(n - 4)}{3(n - 4)} = \dfrac{6(n - 4)}{(n - 8)(n - 4)} $ In order to get the second term over $(n - 8)(n - 4)$ , multiply by $\dfrac{n - 8}{n - 8}$ $ \dfrac{3}{3(n - 4)} \times \dfrac{n - 8}{n - 8} = \dfrac{3(n - 8)}{(n - 8)(n - 4)} $ In order to get the third term over $(n - 8)(n - 4)$ , multiply by $\dfrac{3}{3}$ $ \dfrac{2}{(n - 8)(n - 4)} \times \dfrac{3}{3} = \dfrac{6}{(n - 8)(n - 4)} $ Now we have: $ \dfrac{6(n - 4)}{(n - 8)(n - 4)} - \dfrac{3(n - 8)}{(n - 8)(n - 4)} - \dfrac{6}{(n - 8)(n - 4)} $ $ = \dfrac{ 6(n - 4) - 3(n - 8) - 6} {(n - 8)(n - 4)} $ Expand: $ = \dfrac{6n - 24 - 3n + 24 - 6}{3n^2 - 36n + 96} $ $ = \dfrac{3n - 6}{3n^2 - 36n + 96}$ Simplify: $ = \dfrac{n - 2}{n^2 - 12n + 32}$